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In that article, I prove that the polynomial $X^4+1$ is reducible over all finite prime fields of odd characteristic.

The proof is based on the fact that for $p$ odd prime, the multiplicative group of the non-zero elements of $\mathbb{F}_{p^2}$ has an element of order $8$.

Do you have other proves? Potentially simpler?

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  • $\begingroup$ Well, over a field of characteristic $\;2\;$, the polynomial is reducible: $$x^4+1=(x+1)^4$$ $\endgroup$ – Timbuc Feb 22 '15 at 11:08
  • $\begingroup$ I just payed attention that you seem to state two contradictory things, one in your question's title and the other one in the body of the question. what do you really mean? $\endgroup$ – Timbuc Feb 22 '15 at 11:09
  • $\begingroup$ He can only mean reducible, since irreducible would be false as you pointed out yourself. And for now, we have even two valid proofs that the statement with "reducible" is true. $\endgroup$ – MooS Feb 22 '15 at 11:21
  • $\begingroup$ @Timbuc In the title I mean that the polynomial is irreducible over $\mathbb{Z}$ $\endgroup$ – mathcounterexamples.net Feb 22 '15 at 12:52
  • $\begingroup$ @MooS "...and for now" was way after I wrote the above comment. $\endgroup$ – Timbuc Feb 22 '15 at 12:55
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I think the proof based on $8|p^2-1$ is the simplest proof. Here is another one:

In the algebraic closure $\overline{\mathbb F_p}$ we have the decomposition

$$X^4+1=(X-\frac{1+i}{\sqrt{2}})(X-\frac{1-i}{\sqrt{2}})(X+\frac{1+i}{\sqrt{2}})(X+\frac{1-i}{\sqrt{2}})$$

This gives rise to $3$ compositions in factors of degree $2$:

$X^4+1=(X^2+i)(X^2-i)=(X^2-\sqrt{2}X+1)(X^2+\sqrt{2}X+1)=(X^2-i\sqrt{2}X-1)(X^2+i\sqrt{2}X-1)$

By the well known fact that the product of two non-squares is a square, we deduce that at least one of this decompositions lies in $\mathbb F_p$.

Of course this works for $p$ odd. For $p=2$ the assertion is trivial.

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